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Question 445079: how would you graph the quadratic function y=xsquared-4x+2
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The same way I would graph any quadratic function. Given the quadratic function:
\ =\ ax^2\ +\ bx\ +\ c)
I would first calculate the  -coordinate of the vertex:
Then I would calculate the  -coordinate of the vertex:
\ =\ f\left(\frac{-b}{2a}\right)\ =\ a(x_v)^2\ +\ b(x_v)\ +\ c)
Then I would plot the vertex point: )
Then I would plot the -intercept: )
Then I would use the properties of symmetry to plot the point symmetric to the -intercept: )
Then I would calculate the discriminant,  .
If  then I would know that there are no -intercepts. In this case I would select an integer value close to  and calculate the function value at that independent variable value. I would repeat this step with different independent variable values as often as I felt necessary to give an indication of the shape of the curve.
If  then I would know that I have already plotted the -intercept, namely the vertex. I would proceed to plot another couple points as in the situation described above.
If  , then I would know that there are two distinct real roots and therefore two -intercepts. I would set the function equal to zero and solve the resulting quadratic equation by the most convenient means. Once I had determined the two values for the roots,  and  , I would plot the two points: ) and ) .
Generally speaking, that is sufficient to give an idea of the shape of the curve, so all that is left is to draw a smooth curve through all of your plotted points. If you are unsure of the curve's shape by this point, then by all means select some different values for the independent variable and calculate the function value to give yourself additional data points to use.
John
My calculator said it, I believe it, that settles it
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